編程演算法 - 多重部分和問題 代碼(C)

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多重部分和問題 代碼(C)

本文地址: http://blog.csdn.net/caroline_wendy

題目: 有n種不同大小的數字a, 每種各m個. 判斷是否可以從這些數字之中選出若干使它們的和恰好為K.

使用動態規劃求解(DP), 

方法1: dp[i+1][j] = 用前n種數字是否能加和成j, 時間複雜度O(nKm), 不是最優.

方法2: dp[i+1][j] = 用前i種數加和得到j時, 第i種數最多能剩餘多少個. 時間複雜度O(nK).

例如: n=3, a={3,5,8}, m={3,2,2}, K=17時.

i\j	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17
起始	0	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1	-1
0(3,3)	3	-1	-1	2	-1	-1	1	-1	-1	0	-1	-1	-1	-1	-1	-1	-1	-1
1(5,2)	2	-1	-1	2	-1	1	2	-1	1	2	0	-1	-1	0	1	-1	-1	-1
2(8,2)	2	-1	-1	2	-1	2	2	-1	2	2	2	1	-1	1	1	-1	1	1

代碼:

/* * main.cpp * *  Created on: 2014.7.20 *      Author: spike *//*eclipse cdt, gcc 4.8.1*/#include <stdio.h>#include <memory.h>class Program {static const int MAX_N = 100;int n = 3;int K = 17;int a[MAX_N] = {3,5,8};int m[MAX_N] = {3,2,2};bool dp[MAX_N+1][MAX_N+1];public:void solve() {dp[0][0] = true;for (int i=0; i<n; ++i) {for (int j=0; j<=K; ++j) {for (int k=0; k<=m[i]&&k*a[i]<=j; ++k) {dp [i+1][j] |= dp[i][j-k*a[i]]; //或運算}}}if (dp[n][K]) printf("result = Yes\n");else printf("result = No\n");}};class Program2 {static const int MAX_N = 100;static const int MAX_K = 20;int n = 3;int K = 17;int a[MAX_N] = {3,5,8};int m[MAX_N] = {3,2,2};int dp[MAX_K+1];public:void solve() {memset(dp, -1, sizeof(dp));dp[0] = 0;for (int i=0; i<n; ++i) {for (int j=0; j<=K; ++j) {if (dp[j] >= 0) {dp[j] = m[i];} else if (j < a[i] || dp[j-a[i]]<=0){dp[j] = -1;} else {dp[j] = dp[j-a[i]]-1;}}}if (dp[K]>=0) printf("result = Yes\n");else printf("result = No\n");}};int main(void){Program2 iP;iP.solve();return 0;}

輸出:

result = Yes

編程演算法 - 多重部分和問題 代碼(C)